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• Difficulty Level : Medium
• Last Updated : 31 Aug, 2021

The lower bound for Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick-Sort .. etc) is Ω(nLogn), i.e., they cannot do better than nLogn.

Counting sort is a linear time sorting algorithm that sort in O(n+k) time when elements are in the range from 1 to k.

What if the elements are in the range from 1 to n2
We can’t use counting sort because counting sort will take O(n2) which is worse than comparison-based sorting algorithms. Can we sort such an array in linear time?

Radix Sort is the answer. The idea of Radix Sort is to do digit by digit sort starting from least significant digit to most significant digit. Radix sort uses counting sort as a subroutine to sort.

The Radix Sort Algorithm

1. Do following for each digit i where i varies from least significant digit to the most significant digit.
• Sort input array using counting sort (or any stable sort) according to the i’th digit.

Example:

Original, unsorted list:
170, 45, 75, 90, 802, 24, 2, 66

Sorting by least significant digit (1s place) gives:
[*Notice that we keep 802 before 2, because 802 occurred
before 2 in the original list, and similarly for pairs
170 & 90 and 45 & 75.]

170, 90, 802, 2, 24, 45, 75, 66

Sorting by next digit (10s place) gives:
[*Notice that 802 again comes before 2 as 802 comes before
2 in the previous list.]

802, 2, 24, 45, 66, 170, 75, 90

Sorting by the most significant digit (100s place) gives:
2, 24, 45, 66, 75, 90, 170, 802

What is the running time of Radix Sort?
Let there be d digits in input integers. Radix Sort takes O(d*(n+b)) time where b is the base for representing numbers, for example, for the decimal system, b is 10. What is the value of d? If k is the maximum possible value, then d would be O(logb(k)). So overall time complexity is O((n+b) * logb(k)). Which looks more than the time complexity of comparison-based sorting algorithms for a large k. Let us first limit k. Let k <= nc where c is a constant. In that case, the complexity becomes O(nLogb(n)). But it still doesn’t beat comparison-based sorting algorithms.
What if we make the value of b larger?. What should be the value of b to make the time complexity linear? If we set b as n, we get the time complexity as O(n). In other words, we can sort an array of integers with a range from 1 to nc if the numbers are represented in base n (or every digit takes log2(n) bits).

Applications of Radix Sort :

• In a typical computer, which is a sequential random-access machine, where the records are keyed by multiple fields radix sort is used. For eg., you want to sort on three keys month, day and year. You could compare two records on year, then on a tie on month and finally on the date. Alternatively, sorting the data three times using Radix sort first on the date, then on month, and finally on year could be used.
• It was used in card sorting machines that had 80 columns, and in each column, the machine could punch a hole only in 12 places. The sorter was then programmed to sort the cards, depending upon which place the card had been punched. This was then used by the operator to collect the cards which had the 1st row punched, followed by the 2nd row, and so on.

Is Radix Sort preferable to Comparison based sorting algorithms like Quick-Sort?
If we have log2n bits for every digit, the running time of Radix appears to be better than Quick Sort for a wide range of input numbers. The constant factors hidden in asymptotic notation are higher for Radix Sort and Quick-Sort uses hardware caches more effectively. Also, Radix sort uses counting sort as a subroutine and counting sort takes extra space to sort numbers.

Implementation of Radix Sort

Following is a simple implementation of Radix Sort. For simplicity, the value of d is assumed to be 10. We recommend you to see Counting Sort for details of countSort() function in below code.

## C++

 // C++ implementation of Radix Sort#include using namespace std; // A utility function to get maximum value in arr[]int getMax(int arr[], int n){    int mx = arr[0];    for (int i = 1; i < n; i++)        if (arr[i] > mx)            mx = arr[i];    return mx;} // A function to do counting sort of arr[] according to// the digit represented by exp.void countSort(int arr[], int n, int exp){    int output[n]; // output array    int i, count[10] = { 0 };     // Store count of occurrences in count[]    for (i = 0; i < n; i++)        count[(arr[i] / exp) % 10]++;     // Change count[i] so that count[i] now contains actual    //  position of this digit in output[]    for (i = 1; i < 10; i++)        count[i] += count[i - 1];     // Build the output array    for (i = n - 1; i >= 0; i--) {        output[count[(arr[i] / exp) % 10] - 1] = arr[i];        count[(arr[i] / exp) % 10]--;    }     // Copy the output array to arr[], so that arr[] now    // contains sorted numbers according to current digit    for (i = 0; i < n; i++)        arr[i] = output[i];} // The main function to that sorts arr[] of size n using// Radix Sortvoid radixsort(int arr[], int n){    // Find the maximum number to know number of digits    int m = getMax(arr, n);     // Do counting sort for every digit. Note that instead    // of passing digit number, exp is passed. exp is 10^i    // where i is current digit number    for (int exp = 1; m / exp > 0; exp *= 10)        countSort(arr, n, exp);} // A utility function to print an arrayvoid print(int arr[], int n){    for (int i = 0; i < n; i++)        cout << arr[i] << " ";} // Driver Codeint main(){    int arr[] = { 170, 45, 75, 90, 802, 24, 2, 66 };    int n = sizeof(arr) / sizeof(arr[0]);           // Function Call      radixsort(arr, n);    print(arr, n);    return 0;}

## Java

 // Radix sort Java implementationimport java.io.*;import java.util.*; class Radix {     // A utility function to get maximum value in arr[]    static int getMax(int arr[], int n)    {        int mx = arr[0];        for (int i = 1; i < n; i++)            if (arr[i] > mx)                mx = arr[i];        return mx;    }     // A function to do counting sort of arr[] according to    // the digit represented by exp.    static void countSort(int arr[], int n, int exp)    {        int output[] = new int[n]; // output array        int i;        int count[] = new int[10];        Arrays.fill(count, 0);         // Store count of occurrences in count[]        for (i = 0; i < n; i++)            count[(arr[i] / exp) % 10]++;         // Change count[i] so that count[i] now contains        // actual position of this digit in output[]        for (i = 1; i < 10; i++)            count[i] += count[i - 1];         // Build the output array        for (i = n - 1; i >= 0; i--) {            output[count[(arr[i] / exp) % 10] - 1] = arr[i];            count[(arr[i] / exp) % 10]--;        }         // Copy the output array to arr[], so that arr[] now        // contains sorted numbers according to current digit        for (i = 0; i < n; i++)            arr[i] = output[i];    }     // The main function to that sorts arr[] of size n using    // Radix Sort    static void radixsort(int arr[], int n)    {        // Find the maximum number to know number of digits        int m = getMax(arr, n);         // Do counting sort for every digit. Note that        // instead of passing digit number, exp is passed.        // exp is 10^i where i is current digit number        for (int exp = 1; m / exp > 0; exp *= 10)            countSort(arr, n, exp);    }     // A utility function to print an array    static void print(int arr[], int n)    {        for (int i = 0; i < n; i++)            System.out.print(arr[i] + " ");    }     /*Driver Code*/    public static void main(String[] args)    {        int arr[] = { 170, 45, 75, 90, 802, 24, 2, 66 };        int n = arr.length;                     // Function Call        radixsort(arr, n);        print(arr, n);    }}/* This code is contributed by Devesh Agrawal */

## Python

 # Python program for implementation of Radix Sort# A function to do counting sort of arr[] according to# the digit represented by exp. def countingSort(arr, exp1):     n = len(arr)     # The output array elements that will have sorted arr    output = [0] * (n)     # initialize count array as 0    count = [0] * (10)     # Store count of occurrences in count[]    for i in range(0, n):        index = arr[i] // exp1        count[index % 10] += 1     # Change count[i] so that count[i] now contains actual    # position of this digit in output array    for i in range(1, 10):        count[i] += count[i - 1]     # Build the output array    i = n - 1    while i >= 0:        index = arr[i] // exp1        output[count[index % 10] - 1] = arr[i]        count[index % 10] -= 1        i -= 1     # Copying the output array to arr[],    # so that arr now contains sorted numbers    i = 0    for i in range(0, len(arr)):        arr[i] = output[i] # Method to do Radix Sortdef radixSort(arr):     # Find the maximum number to know number of digits    max1 = max(arr)     # Do counting sort for every digit. Note that instead    # of passing digit number, exp is passed. exp is 10^i    # where i is current digit number    exp = 1    while max1 / exp > 0:        countingSort(arr, exp)        exp *= 10  # Driver codearr = [170, 45, 75, 90, 802, 24, 2, 66] # Function CallradixSort(arr) for i in range(len(arr)):    print(arr[i]) # This code is contributed by Mohit Kumra# Edited by Patrick Gallagher

## C#

 // C# implementation of Radix Sortusing System; class GFG {    public static int getMax(int[] arr, int n)    {        int mx = arr[0];        for (int i = 1; i < n; i++)            if (arr[i] > mx)                mx = arr[i];        return mx;    }     // A function to do counting sort of arr[] according to    // the digit represented by exp.    public static void countSort(int[] arr, int n, int exp)    {        int[] output = new int[n]; // output array        int i;        int[] count = new int[10];         // initializing all elements of count to 0        for (i = 0; i < 10; i++)            count[i] = 0;         // Store count of occurrences in count[]        for (i = 0; i < n; i++)            count[(arr[i] / exp) % 10]++;         // Change count[i] so that count[i] now contains        // actual        //  position of this digit in output[]        for (i = 1; i < 10; i++)            count[i] += count[i - 1];         // Build the output array        for (i = n - 1; i >= 0; i--) {            output[count[(arr[i] / exp) % 10] - 1] = arr[i];            count[(arr[i] / exp) % 10]--;        }         // Copy the output array to arr[], so that arr[] now        // contains sorted numbers according to current        // digit        for (i = 0; i < n; i++)            arr[i] = output[i];    }     // The main function to that sorts arr[] of size n using    // Radix Sort    public static void radixsort(int[] arr, int n)    {        // Find the maximum number to know number of digits        int m = getMax(arr, n);         // Do counting sort for every digit. Note that        // instead of passing digit number, exp is passed.        // exp is 10^i where i is current digit number        for (int exp = 1; m / exp > 0; exp *= 10)            countSort(arr, n, exp);    }     // A utility function to print an array    public static void print(int[] arr, int n)    {        for (int i = 0; i < n; i++)            Console.Write(arr[i] + " ");    }     // Driver Code    public static void Main()    {        int[] arr = { 170, 45, 75, 90, 802, 24, 2, 66 };        int n = arr.Length;         // Function Call        radixsort(arr, n);        print(arr, n);    }     // This code is contributed by DrRoot_}

## PHP

 = 0; \$i--)    {        \$output[\$count[ (\$arr[\$i] /                         \$exp) % 10 ] - 1] = \$arr[\$i];        \$count[ (\$arr[\$i] / \$exp) % 10 ]--;    }     // Copy the output array to arr[], so    // that arr[] now contains sorted numbers    // according to current digit    for (\$i = 0; \$i < \$n; \$i++)        \$arr[\$i] = \$output[\$i];} // The main function to that sorts arr[]// of size n using Radix Sortfunction radixsort(&\$arr, \$n){         // Find the maximum number to know    // number of digits    \$m = max(\$arr);     // Do counting sort for every digit. Note    // that instead of passing digit number,    // exp is passed. exp is 10^i where i is    // current digit number    for (\$exp = 1; \$m / \$exp > 0; \$exp *= 10)        countSort(\$arr, \$n, \$exp);} // A utility function to print an arrayfunction PrintArray(&\$arr,\$n){    for (\$i = 0; \$i < \$n; \$i++)        echo \$arr[\$i] . " ";} // Driver Code\$arr = array(170, 45, 75, 90, 802, 24, 2, 66);\$n = count(\$arr); // Function Callradixsort(\$arr, \$n);PrintArray(\$arr, \$n); // This code is contributed by rathbhupendra?>

## Javascript



Following is the another way of the implementation of the radix sort while using bucket sort technique, it might not look simple while having a look at code but if you give  a shot it’s quite easy, to know more about the bucket sort go here https://www.geeksforgeeks.org/bucket-sort-2 and understand the logic behind the technique.

## C++

Output
6 12 25 161 415 573

Time complexities remains same as in the first method, it’s just the implementation through another method.

Snapshots:

## Quiz on Radix Sort

Other Sorting Algorithms on GeeksforGeeks/GeeksQuiz:

References: